Solve for $x$ and $y$ using substitution. ${x-6y = -3}$ ${x = 5y-2}$
Since $x$ has already been solved for, substitute $5y-2$ for $x$ in the first equation. ${(5y-2)}{- 6y = -3}$ Simplify and solve for $y$ $5y-2 - 6y = -3$ $-y-2 = -3$ $-y-2{+2} = -3{+2}$ $-y = -1$ $\dfrac{-y}{{-1}} = \dfrac{-1}{{-1}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 5y-2}\thinspace$ to find $x$ ${x = 5}{(1)}{ - 2}$ $x = 5 - 2$ ${x = 3}$ You can also plug ${y = 1}$ into $\thinspace {x-6y = -3}\thinspace$ and get the same answer for $x$ : ${x - 6}{(1)}{= -3}$ ${x = 3}$